[Date Index]
[Thread Index]
[Author Index]
Re: Uniform design
*To*: mathgroup at smc.vnet.net
*Subject*: [mg48304] Re: Uniform design
*From*: ab_def at prontomail.com (Maxim)
*Date*: Sat, 22 May 2004 03:04:37 -0400 (EDT)
*References*: <c8kd99$msp$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Bill Rowe <readnewsciv at earthlink.net> wrote in message news:<c8kd99$msp$1 at smc.vnet.net>...
> On 5/20/04 at 4:03 AM, ab_def at prontomail.com (Maxim) wrote:
>
> >Besides, there's a matter of 'optimizations' that Mathematica uses
> >for Sum (an issue which was discussed in this newsgroup a while
> >ago):
>
> >In[1]:= Module[{cnt = 0}, Sum[cnt += 1, {i, 10^6}] ]
> >Module[{cnt =0}, Sum[cnt += 1, {i, 10^6 + 1}] ]
>
> >Out[1]= 500000500000
>
> >Out[2]= 1000001
>
> Clearly, this is a bug. But consider
>
> In[1]:=
> Timing[Module[{cnt = 0}, Sum[cnt += 1, {n, 10^6}]]]
>
> Out[1]=
> {8.69*Second, 500000500000}
>
> In[2]:=
> Timing[Sum[n, {n, 10^6}]]
>
> Out[2]=
> {2.549999999999999*Second, 500000500000}
>
> In[3]:=
> Timing[Sum[n, {n, k}] /. k -> 10^6]
>
> Out[3]=
> {0.019999999999999574*Second, 500000500000}
>
> It seems to me far more natural to due either Sum[n, {n, 10^6}] or Sum[n,
> {n,k}]/.k->10^6 then what you've shown. Not only are these more natural,
> they are faster and avoid the bug.
I'm not so sure about it being a bug; what is happening here is that
for values not greater than 10^6 Sum works as an iterator should
(chapter A.4.2 of the Book), but for larger values Sum calls
SymbolicSum`SymbolicSum, which pre-evaluates the first argument; so
cnt+=1 evaluates to 1 and we're summing 10^6+1 ones. The documentation
says that it is done only for symbolic sums (that is, when there are
non-numerical quantities in the iterator), but my guess is that
Mathematica tries to speed up the computations in this way when the
number of terms is too large. Nobody would want to sum more than 10^6
terms, right?
SymbolicSum`SymbolicSum adds its share of confusion too, because it is
also HoldAll and performs some preliminary operations on the arguments
before evaluating them; in particular, it looks for Out in the first
argument. As a result, we get the following:
In[1]:=
a=k;
Sum[a,{k,n}]
Out[2]=
k*n
In[3]:=
a=k;
Sum[%,{k,n}]
Out[4]=
n*(n+1)/2
That is, there is an incosistency in deciding when the first argument
depends on the iterator variable. And here's another pitfall: we know
that Module[{a=k},Sum[a,{k,n}]] evaluates to k*n, but what if the Sum
remains unevaluated? Mathematica will just return the Sum expression,
but with a already evaluated. So Module[{a=k},Sum[a*f[k],{k,n}]] will
give Sum[k*f[k],{k,n}], not k*Sum[f[k],{k,n}] as in the previous case.
Here's an example:
In[5]:=
Module[{a = 1/k}, Sum[a/k, {k, 1, Infinity}]]
Module[{a = 1/k}, Sum[a/k^2, {k, 1, Infinity}]]
Sum::div: Sum does not converge.
Out[5]=
Pi^2/6
Out[6]=
Pi^2/(6*k)
What's more, Sum[1/k,{k,1,Infinity}] won't necessarily return
unevaluated -- just as well it might evaluate to Infinity, because
there are no clear rules for that either: sometimes Mathematica says
that a sum diverges, sometimes that it converges to Infinity. That
gives yet another possibility for Out[5]: it might return Infinity/k.
So it's anybody's guess what the result will be in such cases.
Maxim Rytin
m.r at prontomail.com
Prev by Date:
**optimally picking one element from each list**
Next by Date:
**Problem with function**
Previous by thread:
**Re: Uniform design**
Next by thread:
**Re: Re: Uniform design**
| |